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of C as a vector space over C, then the list
.1C i; 1 � i/ is linearly dependent.
6 Suppose v1; v2; v3; v4 is linearly independent in V. Prove that the list
v1 � v2; v2 � v3; v3 � v4; v4
is also linearly independent.
7 Prove or give a counterexample: If v1; v2; : : : ; vm is a linearly indepen-
dent list of vectors in V, then
5v1 � 4v2; v2; v3; : : : ; vm
is linearly independent.
8 Prove or give a counterexample: If v1; v2; : : : ; vm is a linearly indepen-
dent list of vectors in V and � 2 F with � ¤ 0, then �v1; �v2; : : : ; �vm
is linearly independent.
9 Prove or give a counterexample: If v1; : : : ; vm and w1; : : : ;wm are lin-
early independent lists of vectors in V, then v1 C w1; : : : ; vm C wm is
linearly independent.
10 Suppose v1; : : : ; vm is linearly independent in V and w 2 V. Prove that
if v1 C w; : : : ; vm C w is linearly dependent, then w 2 span.v1; : : : ; vm/.
38 CHAPTER 2 Finite-Dimensional Vector Spaces
11 Suppose v1; : : : ; vm is linearly independent in V and w 2 V. Show that
v1; : : : ; vm;w is linearly independent if and only if
w … span.v1; : : : ; vm/:
12 Explain why there does not exist a list of six polynomials that is linearly
independent in P4.F/.
13 Explain why no list of four polynomials spans P4.F/.
14 Prove that V is inﬁnite-dimensional if and only if there is a sequence
v1; v2; : : : of vectors in V such that v1; : : : ; vm is linearly independent
for every positive integer m.
15 Prove that F1 is inﬁnite-dimensional.
16 Prove that the real vector space of all continuous real-valued functions
on the interval Œ0; 1� is inﬁnite-dimensional.
17 Suppose p0; p1; : : : ; pm are polynomials in Pm.F/ such that pj .2/ D 0
for each j . Prove that p0; p1; : : : ; pm is not linearly independent in
Pm.F/.
SECTION 2.B Bases 39
2.B Bases
In the last section, we discussed linearly independent lists and spanning lists.
Now we bring these concepts together.
2.27 Deﬁnition basis
A basis of V is a list of vectors in V that is linearly independent and
spans V.
2.28 Example bases
(a) The list .1; 0; : : : ; 0/; .0; 1; 0; : : : ; 0/; : : : ; .0; : : : ; 0; 1/ is a basis of Fn,
called the standard basis of Fn.
(b) The list .1; 2/; .3; 5/ is a basis of F2.
(c) The list .1; 2;�4/; .7;�5; 6/ is linearly independent in F3 but is not a
basis of F3 because it does not span F3.
(d) The list .1; 2/; .3; 5/; .4; 13/ spans F2 but is not a basis of F2 because
it is not linearly independent.
(e) The list .1; 1; 0/; .0; 0; 1/ is a basis of f.x; x; y/ 2 F3 W x; y 2 Fg.
(f) The list .1;�1; 0/; .1; 0;�1/ is a basis of
f.x; y; z/ 2 F3 W x C y C z D 0g:
(g) The list 1; z; : : : ; zm is a basis of Pm.F/.
In addition to the standard basis, Fn has many other bases. For example,
.7; 5/; .�4; 9/ and .1; 2/; .3; 5/ are both bases of F2.
The next result helps explain why bases are useful. Recall that “uniquely”
means “in only one way”.
2.29 Criterion for basis
A list v1; : : : ; vn of vectors in V is a basis of V if and only if every v 2 V
can be written uniquely in the form
2.30 v D a1v1 C � � � C anvn;
where a1; : : : ; an 2 F.
40 CHAPTER 2 Finite-Dimensional Vector Spaces
Proof First suppose that v1; : : : ; vn is a basis of V. Let v 2 V. Because
v1; : : : ; vn spans V, there exist a1; : : : ; an 2 F such that 2.30 holds. To
This proof is essentially a repeti-
tion of the ideas that led us to the
deﬁnition of linear independence.
show that the representation in 2.30 is
unique, suppose c1; : : : ; cn are scalars
such that we also have
v D c1v1 C � � � C cnvn:
Subtracting the last equation from 2.30, we get
0 D .a1 � c1/v1 C � � � C .an � cn/vn:
This implies that each aj � cj equals 0 (because v1; : : : ; vn is linearly inde-
pendent). Hence a1 D c1; : : : ; an D cn. We have the desired uniqueness,
completing the proof in one direction.
For the other direction, suppose every v 2 V can be written uniquely in
the form given by 2.30. Clearly this implies that v1; : : : ; vn spans V. To show
that v1; : : : ; vn is linearly independent, suppose a1; : : : ; an 2 F are such that
0 D a1v1 C � � � C anvn:
The uniqueness of the representation 2.30 (taking v D 0) now implies that
a1 D � � � D an D 0. Thus v1; : : : ; vn is linearly independent and hence is a
basis of V.
A spanning list in a vector space may not be a basis because it is not
linearly independent. Our next result says that given any spanning list, some
(possibly none) of the vectors in it can be discarded so that the remaining list
is linearly independent and still spans the vector space.
As an example in the vector space F2, if the procedure in the proof below
is applied to the list .1; 2/; .3; 6/; .4; 7/; .5; 9/, then the second and fourth
vectors will be removed. This leaves .1; 2/; .4; 7/, which is a basis of F2.
2.31 Spanning list contains a basis
Every spanning list in a vector space can be reduced to a basis of the
vector space.
Proof Suppose v1; : : : ; vn spans V. We want to remove some of the vectors
from v1; : : : ; vn so that the remaining vectors form a basis of V. We do this
through the multi-step process described below.
Start with B equal to the list v1; : : : ; vn.
SECTION 2.B Bases 41
Step 1
If v1 D 0, delete v1 from B . If v1 ¤ 0, leave B unchanged.
Step j
If vj is in span.v1; : : : ; vj�1/, delete vj from B . If vj is not in
span.v1; : : : ; vj�1/, leave B unchanged.
Stop the process after step n, getting a list B . This list B spans V because our
original list spanned V and we have discarded only vectors that were already
in the span of the previous vectors. The process ensures that no vector in B
is in the span of the previous ones. Thus B is linearly independent, by the
Linear Dependence Lemma (2.21). Hence B is a basis of V.
Our next result, an easy corollary of the previous result, tells us that every
ﬁnite-dimensional vector space has a basis.
2.32 Basis of ﬁnite-dimensional vector space
Every ﬁnite-dimensional vector space has a basis.
Proof By deﬁnition, a ﬁnite-dimensional vector space has a spanning list.
The previous result tells us that each spanning list can be reduced to a basis.
Our next result is in some sense a dual of 2.31, which said that every
spanning list can be reduced to a basis. Now we show that given any linearly
possibility of adjoining no additional vectors) so that the extended list is still
linearly independent but also spans the space.
2.33 Linearly independent list extends to a basis
Every linearly independent list of vectors in a ﬁnite-dimensional vector
space can be extended to a basis of the vector space.
Proof Suppose u1; : : : ; um is linearly independent in a ﬁnite-dimensional
vector space V. Let w1; : : : ;wn be a basis of V. Thus the list
u1; : : : ; um;w1; : : : ;wn
spans V. Applying the procedure of the proof of 2.31 to reduce this list to a
basis of V produces a basis consisting of the vectors u1; : : : ; um (none of the
u’s get deleted in this procedure because u1; : : : ; um is linearly independent)
and some of the w’s.
42 CHAPTER 2 Finite-Dimensional Vector Spaces
As an example in F3, suppose we start with the linearly independent
list .2; 3; 4/; .9; 6; 8/. If we take w1;w2;w3 in the proof above to be the
standard basis of F3, then the procedure in the proof above produces the list
.2; 3; 4/; .9; 6; 8/; .0; 1; 0/, which is a basis of F3.
Using the same basic ideas but
Suppose